Exercise: Continuity and Local Boundedness Consider function $f:\mathbb{R}^n \setminus \{0\} \times \mathbb{R}^n \rightarrow \mathbb{R}_{> 0}$ that is:


*

*continuous in the first argument;

*locally bounded in the second argument.
Consider a sequence $\{x_i\}_{i=1}^{\infty}$ such that $x_i \in \mathbb{R}^n$, $x_i \rightarrow x$.
Prove that
$$ \limsup_{i \rightarrow \infty} \ f(x,x_i) - f(x_i,x_i) \ = \ 0. $$
 A: I believe the following is a counter-example.
To keep things simple, let $n=1$. Consider the sequence $x_n = 1-\frac{1}{2^n}$. This sequence converges to $x=1$. Now define a function $f:\mathbb{R}\setminus\lbrace0\rbrace\times\mathbb{R}\to\mathbb R_{>0}$ as follows: $$f(t,y):=\begin{cases}1&&\textrm{if }y\geq1\\1 && \textrm{if }t\leq y<1\\1+\frac{t-x_n}{x-x_n}&&\textrm{if }y<t<1\\2&&\textrm{if }y<1\leq t\end{cases}$$
This function is bounded (it takes values between $1$ and $2$), so it is also locally bounded in the second argument. It is also continuous in the first argument, since it is piecewise linear for each fixed $y$ and the boundary values (or limits) agree.
But, $f(x,x_n)-f(x_n,x_n)=2-1=1$ for each $n\in\mathbb{N}$, so $\lim_{n\to\infty}(f(x,x_n)-f(x_n,x_n))=1$, which means $\limsup$ is also equal to $1\neq0$.
In case you want a counter-example for $n\in\mathbb{N}$, a similar construction should work. The point here is that you can define a continuous function $f_y:\mathbb{R}^n\setminus\lbrace0\rbrace\to\mathbb{R}_{>0}$ pretty much independently for each $y\in\mathbb R^n$ and then just put them together by $f(t,y) = f_y(t)$ as long as you make sure they stay sufficiently bounded. This independence can then enable some pretty wild behaviour.
